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the correct answer will be marked as the brainlest. hey, guys please do me a favor and check my answer for where am I wrong the Q is " From a point on the ground 40m away from the foot of the tower, the angle of elevation of the top of the tower is 30. The angle of elevation of the top of the water tanking the top of the tower is 45, find the height of the tower and the depth of the tank. " MY ANSWER- taking ΔDCB With ∠DCB=30° CB=40m DB÷40=1/2 (Sin 30° {Sin C =AB/AC}) DB=20⇒ [EQ. 1] In ΔABC [ note:AD is the tank above the tower ] ∠ACB = 45° CB=40m AB/BC=Sin 45° AB= 40/√2⇒ (40√2) /2---------EQ. 2 AD= AB-BD AD= (40√2)/2 - 20 = ⇒ 20√2-20= √2 Hence, AD=√2 PLEASE TELL ME WHERE AM I WRONG?

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thupirocks2548

08.01.2018

Math

Secondary School

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From a point on the ground 40m away from the foot of the tower ,the angle of elevation of the top of the tower is 30. The angle of elevation of the top of the water tanking the top of the tower is 45, find height of the tower and the depth of the tank

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alessre

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Hello,

let h m be the height of tower and m be the depth of tank (as figure).

Now, in ΔABC:

tan30°=BC/AB;

1/√3=h/40;

h=40/√3 m (1)

In ΔABD:

tan45°=DC+BC/AB;

1=x+h/40;

x+h=40;

using (1):

x+40/√3=40;

x=40(1-1/√3) m

Then height of the tower = 40/√3 m,

and depth up tank :

40(1-1/√3)m

bye :-)